I am trying to understand the proof that the axis parallel rectangles are PAC learnable in the realizable case. This means that given $\epsilon, \delta$ with enough data we can find a function $h$ such that $$\mathbb{P}\left[\text{error} > \epsilon\right] \leq \delta$$ Here error can be seen as the probability of making an error with our chosen function $h$.
Now, for axis parallel rectangles (in binary classification) the usual argument goes like this, let $R$ be the true rectangle and let $R'$ be the smallest rectangle containing the positive examples, clearly $R' \subseteq R$, we consider the four rectangular strips between $R'$ and $R$. Clearly if all of them have probability $\leq \epsilon/4$ then the probability of making an error is less than $\epsilon$, so we can assume that at least one have probability of making an error $\geq \epsilon/4$.
For such a strip the probability of correctly classifying all the $m$ training examples is at most $(1 - \epsilon/4)^m$, and thus taking aunion bound over all strips we get that the probability of correctly classifying everything is less than $4(1-\epsilon/4)^m \leq 4e^{-m/4}$, and with a bit of algebra this yields that the sample complexity is $m \geq (4/\epsilon)\ln(4/\delta)$.
Here is a pdf that explains in a bit more detail, with some pictures, I just had to condense the argument as much as I can to fit it in here.
My question is, why do we have to consider the four rectangular strips separatedly, why can't we just say that the probability of the region between $R'$ and $R$ has to be greater than $\epsilon$ (because otherwise we are done), and thus using the same argument we would arrive at the better bound $m \geq (1/\epsilon)\ln(1/\delta)$?
Sorry for the long question, and thanks in advance.